Do 2kg wody o temperaturze 10*C wrzucono 100g żalaza o temperaturze 80*C. Jaka będzie temperatura końcowa?

m₁ = 2 kg T₁ = 10 C cw = 4200 J/kg * C   m₂ = 100 g = 0,1 kg T₂ = 80 C cw = 460 J/kg * C T = ? Q₁ = Q₂ [latex]m_1 * c_w_1* ( T - T_1 ) = m_2 * c_w_2* ( T_2 - T )[/latex] [latex]2 kg * 4200 frac{J}{kg * C } * (T - 10 C)= 0,1 kg * 460 frac{J}{kg * C * (80 C - T )} [/latex] [latex]8400 frac{J}{C} ( T-10 C ) = 46 frac{J}{C} * ( 80 C - T) 8400 J/C * T - 84000 J = 3680 J - 46 J/C * T 8400 J/C * T + 46 J/C * T = 3680 J + 84000 J 8446 J/C * T = 87680 J T = 10,38 C[/latex]